数理解析研究所講究録
1080 巻 1999 年 34-40
34
Equivalence relation between an order preserving
operator inequality and related operator functions
東京理科大
東京理科大
東京理科大
理
理
理
伊藤
橋本
古田
滴瓶
雅史
孝之
(Masatoshi Ito)
(Masashi Hashimoto)
(Takayuki Furuta)
Abstract
This report is based on the following two papers:
[FHI] T.Furuta, M.Hashimoto and M.Ito, Equivalence relation between generalized
Furuta inequality and related operator functions, Scientiae Mathematicae, 1
(1998), 257-259.
.
[F] T.Furuta, Simplified proof of an order preserving operator $inequali\dot{t}y$ , Proc.
Japan. Acad., 74 (1998), 114.
In this paper, we shall show equivalence relation between an order preserving
operator inequality and related operator functions.
1
Introduction
Chapter 1, 2 and 3 are based on [FHI].
A capital letter means a bounded linear operator on a complex Hilbert space $H$ . An
is said to be positive (denoted by $T\geq 0$ ) if $(Tx, x)\geq 0$ for all $x\in H$ and
operator
is positive
is said to be strictly positive (denoted by $T>0$ ) if
also an operator
L\"owner-Heinz
and invertible. The following Theorem is an extension of the celebrated
theorem: $A\geq B\geq 0$ ensures
for any $\alpha\in[0,1]$ .
$T$
$T$
$T$
$\mathrm{F}$
$A^{\alpha}\geq B^{\alpha}$
Theorem
$\mathrm{F}([5])$
If $A\geq B\geq 0$ ,
(i)
.
then for each
$r\geq 0$
,
$(B^{\frac{r}{2}}A^{p}B \frac{r}{2})^{\frac{1}{q}}\geq(B^{\frac{r}{2}}B^{p}B\frac{r}{2})^{\frac{1}{q}}$
and
(ii)
$(A^{\frac{r}{2}}A^{p}A \frac{r}{2})^{\frac{1}{q}}\geq(A^{\frac{r}{2}}B^{p}A\frac{r}{2})^{\frac{1}{q}}$
hold for $p\geq 0$ and $q\geq 1$ with
$(1+r)q\geq p+r$ .
We remark that Theorem yields L\"owner-Heinz theorem when we put $r=0$ in (i)
or (ii) stated above. Alternative proofs of Theorem are given in $[2][14]$ and also an
and
elementary one-page proof in [6]. It is shown in [15] that the domain drawn for
in the Figure is best possible one for Theorem F.
In [9] we established the following Theorem as extensions of Theorem F.
$\mathrm{F}$
$\mathrm{F}$
$p,$
$\mathrm{G}$
$q$
$r$
35
Theorem
$\mathrm{G}([9])$
. If
$A\geq B\geq 0$
with $A>0$ , then for each
and $p\geq 1$ ,
$t\in[0,1]$
$F_{p,t}(A, B, r, S)=A^{\frac{-r}{2}\{}A \frac{r}{2}(A\frac{-t}{2}B^{p}A^{\frac{-t}{2})}sA^{\frac{r}{2}}\}^{\frac{1-t+r}{(p-t)s+f}}A^{\frac{-r}{2}}$
is decreasing for $r\geq t$ and
$t\in[0,1]$ and $p\geq 1$ ,
$s\geq 1_{f}$
and
$F_{p,t}(A, A, r, S)\geq F_{p,t}(A, B, r, S)_{f}$
that is,
for
each
$A^{1t+r}- \geq\{A^{\frac{r}{2}}(A\frac{-t}{2}BpA^{\frac{-t}{2})^{s}A^{\frac{r}{2}\}^{\frac{1-t+r}{(p-t)s+r}}}}$
holds for any
and
$s\geq 1$
$r\geq t$
.
established excellent
useful inequality equivalent to the main
If $A\geq B\geq 0$ with $A>0$ , then
$\log$
$\mathrm{A}\mathrm{n}\mathrm{d}\mathrm{o}-\mathrm{H}\mathrm{i}\mathrm{a}\mathrm{i}[1]$
majorization results and proved the following
theorem: .
$\log \mathrm{m}\mathrm{a}\mathrm{j}_{\mathrm{o}\mathrm{r}}\mathrm{i}\mathrm{Z}\mathrm{a}\mathrm{t}\mathrm{i}_{\mathrm{o}\mathrm{n}}\downarrow$
$A^{r} \geq\{A^{\frac{r}{2}}(A^{\frac{-1}{2}}B^{p}A^{\frac{-1}{2})^{r}A}\frac{r}{2}\}^{\frac{1}{p}}$
holds for any $p\geq 1$ and $r\geq 1$ .
Theorem interpolates the inequality stated above by Ando-Hiai and Theorem itself,
and also extends results of $[3][7]$ and [8]. Recently a nice mean theoretic proof of Theorem
is shown in [4] and the result on the best possibility of Theorem is shown in [16].
Very recently the following Theorem is obtained in [10] as an extension of Theorem G.
$\mathrm{F}$
$\mathrm{G}$
$\mathrm{G}$
$\mathrm{G}$
$\mathrm{H}$
Theorem
$\mathrm{H}([10])$
$p \geq\max\{q, t\}$
.
Let
$A\geq B\geq 0$
with $A>0$ . For each
$t\in[0,1],$
$q\geq 0$
and
,
$G_{p,q,t}(A, B, r, s)=A^{\frac{-r}{2}\{}A \frac{r}{2}(A^{\frac{-t}{2}}BpA^{\frac{-t}{2})}sA^{\frac{r}{2}}\}^{\frac{q-t+r}{(p-t)s+f}}A^{\frac{-r}{2}}$
. Moreover for each
$G_{p,q,t}(A, A, r, S)\geq G_{p,q,t}(A, B, r, s)_{y}$ that is,
is decreasing for
$r\geq t$
and
$s\geq 1$
$1\geq q\geq t\geq 0$
and $p\geq q$ ,
$A^{q-t+r}\geq\{A^{\frac{r}{2}}(A^{\frac{-\mathrm{t}}{2}}B^{p}A^{\frac{-t}{2})^{s}A^{\frac{r}{2}\}^{\frac{q-t+r}{(_{P^{-t}})S+\Gamma}}}}$
holds for any
$\mathrm{H}$
$r\geq t$
and
Alternative proofs of
is extended in [11].
$s\geq 1$
.
$\overline{\mathrm{T}}\mathrm{h}\mathrm{e}\mathrm{o}\mathrm{r}\mathrm{e}\mathrm{m}\mathrm{H}$
are. shown in
$[12][13]$ ,
and
th.
$\mathrm{e}\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{d}\mathrm{i}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}$
.
of Theorem
Result
2
We obtain the following Theorem 1 associated with Theorem
$\mathrm{G}$
and Theorem H.
Theorem 1. The following statements hold and follow from each other;
(i)
If $A\geq B\geq 0$
with $A>0$ , then for each
$t\in[0,1]$
and
$p\geq 1_{f}$
$A^{1-t+r}\geq\{A^{\frac{r}{2}}(A^{\frac{-t}{2}}BpA^{\frac{-t}{2})^{S}}A^{\frac{r}{2}}\}^{\frac{1-l+r}{(\mathrm{p}-t)_{S}+\Gamma}}$
holds for any
$s\geq 1$
and $r\geq t$ .
36
(ii)
If $A\geq B\geq 0$
with
$A>0_{f}$
then for each
$1\geq q\geq t\geq 0$
and $p\geq q$ ,
$A^{q-t+r} \geq\{A^{\frac{r}{2}}(A^{\frac{-t}{2}}B^{p}A^{\frac{-t}{2})^{s_{A\}^{\frac{q-\mathrm{t}+r}{(p-t)s+r}}}}}\frac{r}{2}$
(iii)
holds for any
$s\geq 1$
If $A\geq B\geq 0$
with
and
$r\geq t$
$A>0_{y}$
.
then for each
$t\in[0,1]$
and $p\geq 1$ ,
$F_{p,t}(A, B, r, S)=A^{\frac{-r}{2}\{A^{\frac{r}{2}}(A^{\frac{-t}{2}BA^{\frac{-\mathrm{t}}{2})}}}pSA \frac{r}{2}\}^{\frac{1-t+r}{(\mathrm{p}-t)s+r}}A^{\frac{-f}{2}}$
is decreasing for
(iv)
If $A\geq B\geq 0$
and
$r\geq t$
$s\geq 1$
.
with $A>0$ , then for each
$t\in[0,1],$
$q\geq 0$
and $p\geq t$ ,
$G_{p,q,t}(A, B, r, S)=A^{\frac{-r}{2}\{A^{\frac{f}{2}}(A^{\frac{-t}{2}BA^{\frac{-t}{2})}}}pSA \frac{r}{2}\}^{\frac{q^{-t}+\Gamma}{(p-t)s+r}}A^{\frac{-r}{2}}$
is decreasing for
and
$r\geq t$
$s\geq 1$
such that $(p-t)s\geq q-t$ .
(i) and (iii) in Theorem 1 have been obtained as Theorem G. (ii) and (iv) have been also
obtained as Theorem and an extension of Theorem H. (ii) and (iv) used to be recognized
extensions of (i) and (iii) respectively, and (i), (ii), (iii) and (iv) proved sepalately. Theorem
1 asserts that (i), (ii), (iii) and (iv) follow from each other. In other words, if we prove
only one of them, then we obtain the others. So, in this report, we shall give a simplified
proof of (ii), too.
We need the following lemma to give proofs.
$\mathrm{H}$
Lemma F. Let $A>0$ and
$B$
be an invertible operator. Then
$(BAB^{*}) \lambda--BA\frac{1}{2}(A^{\frac{1}{2}}B*BA\frac{1}{2})^{\lambda}-1A^{\frac{1}{2}}B*$
holds for any real number .
$\lambda$
3
Proof of Theorem 1
We may assume that
is invertible without loss of generality. (i), (ii), (iii) and (iv)
have been already proved in [9] $[10][11][12]$ and [13], so that we have only to prove the
equivalence relation among them. We shall show
as follows.
$\mathrm{B}$
$(\mathrm{i}\mathrm{v})arrow(\mathrm{i}\mathrm{i}\mathrm{i})arrow(\mathrm{i})arrow(\mathrm{i}\mathrm{i})arrow(\mathrm{i}\mathrm{v})$
Proof of
Proof of
. Put
and let $p\geq 1$ in (iv), we have (iii).
. In (iii), comparing the functional value and the maximal value of
$F_{p,t}(A, B, \Gamma, S)$ , we have the following inequality: For each $t\in[0,1]$ and $p\geq 1$ ,
$(\mathrm{i}\mathrm{v})\Rightarrow(\mathrm{i}\mathrm{i}\mathrm{i})$
$q=1$
1
$(\mathrm{i}\mathrm{i}\mathrm{i})\Rightarrow(\mathrm{i})$
$F_{p,t}(A, B, r, S)\leq F_{p,t}(A, B,t, 1)$
(3.1)
$=A^{\frac{-t}{2}BA^{\frac{-t}{2}}}$
$\leq A^{1-t}$
holds for any
$r\geq t$
and
$s\geq 1$
. Multiplying
$A^{\frac{r}{2}}$
by
$A\geq B$
from the both sides of (3.1), we have (i).
37
Proof of
$(\mathrm{i})\Rightarrow(\mathrm{i}\mathrm{i})$
theorem. Put
. Since
$q\in[0,1],$ $A\geq B\geq 0$
$A_{1}=A^{q},$ $B_{1}=B^{q},$
assures $A^{q}\geq B^{q}$ by L\"owner-Heinz
and
in (i). Then
$t_{1}=arrow q\theta\in[0,1],$ $p_{1}= \frac{p}{q}\geq 1$
$A_{1}^{1-t_{1}+1}r=A^{q-t+r},$
$r_{1}= \frac{r}{q}\geq t_{1}$
$A_{1^{-\lrcorner}}^{-_{2}}=\iota A^{\frac{-t}{2}},$
$A_{1}^{\lrcorner}r_{2}=A^{\frac{r}{2}},$
$B_{1}^{p_{1}}=B^{p}$
and
$\underline{1-t_{1}+r_{1}}\underline{q-t+\Gamma}=$
$(p_{1}-t_{1})_{S}$
\dagger
$(p-t)s+r$ ’
$r_{1}$
that is, we have the following (3.2): For each
$1\geq q\geq t\geq 0$
and
$p\geq q$
,
(3.2)
$A^{q-t+r} \geq\{A^{\frac{r}{2}}(A^{\frac{-t}{2}}B^{p}A^{\frac{-t}{2})^{s_{A\}^{\frac{q-t+r}{(p-t)S+r}}}}}\frac{r}{2}$
holds for any
Proof of
For each
and $s\geq 1$ . Hence the proof of
is complete.
. Put
and $q=t$ in (ii), we hava the following (3.3):
and $p\geq t$ ,
$r\geq t$
$(\mathrm{i})\Rightarrow(\mathrm{i}\mathrm{i})$
$D=A^{\frac{-t}{2}B^{p}A^{\frac{-t}{2}}}$
$(\mathrm{i}\mathrm{i})\Rightarrow(\mathrm{i}\mathrm{v})$
$t\in[.0,1]$
(3.3)
$A^{r}\geq(A^{\frac{r}{2}}D^{s}A^{\frac{r}{2}})^{\frac{r}{\{\mathrm{p}-t)s+r}}$
holds for any
$r\geq t$
and
$s\geq 1$
.
$(3.3)$
(
is equivalent to the following (3.4) by Lemma F.
$D^{\frac{\mathrm{e}}{2}}.A^{r}D^{\frac{s}{2})^{\frac{(p-t)}{(\mathrm{p}-t)_{S}+r}}}..\geq D^{s}$
.
(3.4)
Applying L\"owner-Heinz theorem to (3.3) and (3.4), we have the following (3.5) and (3.6):
$A^{u}\geq(A^{\frac{r}{2}}D^{s}A^{\frac{r}{2})^{\frac{u}{(p-t)\underline{\mathrm{q}}+r}}}$
(
$D^{\frac{s}{2}}A^{r}D^{\frac{\epsilon}{2})^{\frac{(p-t)w}{(_{P^{-t}})\vee r\mathrm{s}+}}}\geq D^{w}$
for
$r\geq u\geq 0$
for
$s\geq w\geq 0$
.
(3.5)
.
In order to show that $G_{p,q,t}(A, B, r, S)$ is a decreasing function of both
$G_{p,q,t}(A, B, r, S)$ as follows:
$G_{p,q,t}(A, B, r, s)=$
$=$
$=$
$=$
(3.6)
$r$
and , we rewrite
$s$
$A^{\frac{-r}{2}(A)^{\frac{-t+r}{(\mathrm{p}-t)S+r}}A}A^{\frac{r}{2}}D^{s} \frac{r}{2}\frac{-r}{2}$
$A^{\frac{-r}{2}f(S})A^{\frac{-r}{2}}$
$D^{\frac{s}{2}}(D^{\frac{s}{2}}A^{r}D \frac{s}{2})^{\frac{q-t-(prightarrow t)\underline{\mathrm{Q}}}{(p-t)s+\Gamma}}D^{\frac{s}{2}}$
by Lemma
$\mathrm{F}$
$D^{\frac{s}{2}}g(r)D^{\frac{s}{2}}$
where
$f(s)\equiv(A^{\frac{r}{2}}D^{s_{A^{\frac{r}{2})^{\frac{q-t+r}{(p-t)s+\Gamma}}}}}$
and
$g(r)\equiv(D^{\frac{s}{2}}Ar_{D^{\frac{s}{2})^{\frac{9^{-t-}\mathrm{t}P-t)S}{(p-t)S+r}}}}$
So we have only to prove that $f(s)$ and $g(r)$ are decreasing for and respectively.
(a) Proof of the result that $G_{p,q,t}(A, B, r, S)$ is decreasing for $s\geq 1$ such that $(p-t)s\geq
$r$
$f(s)$
$=$
$=$
$=$
$\geq$
$=$
$=$
$s$
$(A^{\frac{r}{2}}D^{S}A^{\frac{r}{2}})^{\frac{-t+r}{(p-t)\vee\Gamma 8+}}$
$\{(A^{\frac{r}{2}}DSA\frac{r}{2})^{\frac{(p^{-\mathrm{t}})(\mathrm{e}+w)+r}{(\mathrm{p}-t)\mathrm{q}+r}}..\}^{\frac{q-t+r}{(p-t)(s+w)+r}}$
$\{A^{\frac{r}{2}}D^{\frac{s}{2}}(D\frac{s}{2}A^{r}D\frac{s}{2})^{\frac{(p-t)w}{(p-\mathrm{t})s+f}}D\frac{s}{2}A^{\frac{r}{2}}\}\frac{q-t+r}{(p-t)(_{S+}w)+r}$
for $s\geq w\geq 0$
by Lemma
$\mathrm{F}$
$(A^{\frac{r}{2}}D^{\frac{s}{2}}D^{w}D^{\frac{s}{2}}A^{\frac{r}{2}})^{\frac{-t+r}{(p-t)(_{8+}w)+r}}$
$(A^{\frac{r}{2}}D^{s+w}A^{\frac{r}{2}})^{\frac{-t+r}{(p-t)1S+w)+r}}$
$f(s+w)$ .
q-t$ .
38
The last inequality holds by (3.6) and L\"owner-Heinz theorem since
holds by the condition of (ii). Hence
$s\geq 1$ such that $(p-t)s\geq q-t$ .
(b) Proof of the result that $G_{p,q,t}(A, B, r, s)$ is decreasing for $r\geq t$ .
$\frac{q-t+r}{(p-t)(S+w)+r}\in[0,1]$
$G_{p,q,t}(A, B, \Gamma, S)=A^{\frac{-r}{2}f(S})A^{\frac{-r}{2}}$
$g(r)$
is decreasing for
$D^{\frac{s}{2}}(D^{\frac{l}{2}}.A^{\cdot} \Gamma D\frac{s}{2})^{\frac{q-t-\{p-t)\mathrm{q}}{(\mathrm{p}-t)\mathrm{s}+r}}.\cdot D^{\frac{s}{2}}$
$=$
$D^{\frac{s}{2}} \{(D^{\frac{s}{2}A^{\Gamma}}D^{\frac{s}{2}})\frac{(\mathrm{p}-t)\underline{9}+r+u}{(p-t)\mathit{8}+r}\}^{\frac{q-t-(p-t)_{\vee}\mathrm{q}}{(p-t)\mathrm{S}+r+u}}D\frac{s}{2}$
$=$
$D^{\frac{s}{2}} \{D^{\frac{s}{2}}A\frac{r}{2}(A\frac{f}{2}D^{S}A\frac{r}{2})^{\frac{u}{(\mathrm{p}-t)s+r}}A^{\frac{r}{2}}D^{\frac{\epsilon}{2}\}^{\frac{q-t-(p-t)*}{1^{p-t})S+r+u}}}D^{\frac{s}{2}}$
$=$
for $r\geq u\geq 0$
by Lemma
$\mathrm{F}$
$D^{\frac{s}{2}} \{D^{\frac{s}{2}}A\frac{r}{2}A^{u}A^{\frac{r}{2}}D\overline{2}\}\vee^{*}\frac{q-t-(\mathrm{p}-t)*}{(p-t)8+r+u}.D^{\frac{s}{2}}$
$\underline{>}$
$D^{\frac{s}{2}}(D^{\frac{s}{2}}A^{r+u}D^{\frac{\epsilon}{2}})^{\frac{q^{-t-}(p-t)\mathrm{Q}}{(\mathrm{p}-t)\mathrm{s}+r+u}}.D^{\frac{s}{2}}$
$=$
$=g(r+u)$ .
The last inequality holds by (3.5), L\"owner-Heinz theorem and taking inverses since
by the condition of (ii). Hence
$G_{p,q,t}(A, B, r, S)=D^{\frac{\sigma}{2}}g(r)D^{*}\overline{2}$
is
$\frac{q^{-t-}(p-t)s}{\mathrm{s}_{\mathrm{e}\mathrm{C}}p-t\mathrm{r}\mathrm{e})s+r+\mathrm{a}\sin^{u}}\in[-1,0]\mathrm{h}_{\mathrm{o}\mathrm{l}\mathrm{d}\mathrm{S}}\mathrm{g}\mathrm{f}\mathrm{o}\mathrm{r}r\geq t$
.
is complete by (a) and (b).
Consequently we have finished the proof of Theorem 1.
So the proof of
$(\mathrm{i}\mathrm{i})\Rightarrow(\mathrm{i}\mathrm{v})$
Simplified proof of (ii)
4
This chapter is based on [F].
(ii) in Theorem 1. Firstly we shall show the following (4.1) which is the case
of $r=t$ in (ii): If $A\geq B\geq 0$ with $A>0$ , then
Proof of
(4.1)
$A^{q} \geq\{A^{\frac{t}{2}}(A^{\frac{-t}{2}B^{p}A}\frac{-t}{2})sA\frac{t}{2}\}^{\frac{q}{(p-t)s+t}}$
holds for $1\geq q\geq t\geq 0,$ $p\geq q$ and $s\geq 1$ .
and
(1st) In case $2\geq s\geq 1$ . $s-1\in[0,1],$
condition of (ii) and L\"owner-Heinz theorem. Then we have
$\frac{q}{(p-t)_{S+t}}\in[0,1]$
$B_{1}$
$=$
$=$
$\leq$
$=$
for
$A^{t}\geq B^{t}\cdots(**)$
hold by the
$\{A^{\frac{t}{2}(A^{\frac{-t}{2}B^{p}A}}\frac{-t}{2})sA^{\frac{t}{2}}\}(\mathrm{p}-\sim_{s})t\overline{+t}$
$\{B^{E}2(B^{\epsilon}2A^{-}tB^{f}2\mathrm{i})^{s-}1B2\}^{\frac{q}{(\mathrm{p}-t)\epsilon+t}}\epsilon$
by Lemma
$\{B^{E}2(B^{E}2B-tB^{\mathrm{s}\mathrm{i}}2)s-1B2E\}^{\overline{(})_{S}t}p-t\mapsto\overline{+}$
by
$\mathrm{F}$
$(**)$
(4.2)
$B^{q}\leq A^{q}=A_{1}$
and $2\geq s\geq 1$ .
(2nd) Repeating (4.2) for $A_{1}\geq B_{1}\geq 0$ , we obtain the following (4.3):
$1\geq q\geq t\geq 0,$ $p\geq q$
$A_{1}^{q_{1}} \geq\{A_{1}^{2}(A^{-}2B_{1}^{p}1A^{\frac{-t_{\rceil}}{12}}\underline{t}_{1}1^{-\perp\perp}\iota\underline{t})S_{1}A_{1}^{2}\}\frac{q\rceil}{(_{P1^{-t_{1}}})\epsilon 1+t1}$
holds for
12
$q_{1}\geq t_{1}\geq 0,$ $p_{1}\geq q_{1}$
and
$2\geq s_{1}\geq 1$
.
(4.3)
39
By putting $1=q_{1} \geq t_{1}=\frac{t}{q}\geq 0,$
we obtain the following (4.4):
$p_{1}= \frac{\langle p-t)S+t}{q}\geq q_{1}=1$
and restoring
$A_{1}$
and
$B_{1}$
in (4.3),
(4.4)
$A^{q} \geq\{A^{\frac{t}{2}}(A^{\frac{-t}{2}}B^{p}A^{\frac{-t}{2})A}ss1\frac{t}{2}\}^{\infty_{Ss}}\mathrm{t}p-\mathrm{t})1+t$
holds for $1\geq q\geq t\geq 0,.p\geq q$ and
(4.4), (4.1) holds for any $s\geq 1$ .
$4\geq ss_{1}\geq 1$
.
$\mathrm{B}\mathrm{y}$
repeating this process from (4.2) to
,
Lastly let
$A_{2}=A^{q}$
Then (4.1) assures
and
$A_{2}\geq B_{2}\geq 0$
$B_{2}= \{A^{\frac{t}{2}}(A^{\frac{-t}{2}B^{p}}A^{\frac{-t}{2})\}}SA\frac{t}{2}\overline{(p-}t)s\mathrm{L}\overline{+t}$
. By Theorem
.
$\mathrm{F}$
(4.5)
$A_{2}1+r_{2}\geq(A_{2}B_{2^{p2}}A2)^{p+r}\underline{r}_{2}zrZarrow 21+r22$
holds for $p_{2}>1$ and
Put
following (4.6): For each
$r_{2}\geq 0$
$p_{2}= \frac{(p-\overline{t})S+t}{q}\geq 1,$
.
and restore
$1\geq q\geq t\geq 0$ and $p\geq q$ ,
$A_{2}$
$r_{2}= \frac{r-t}{q}\geq 0$
and
$B_{2}$
in (4.5), then we obtain the
(4.6)
$A^{q-t+r} \geq\{A\frac{r}{2}(A\frac{-t}{2}BpA^{\frac{-t}{2})^{s_{A\}^{\frac{q-t+r}{(p-t)s+r}}}}}\frac{r}{2}$
holds for any $r\geq t$ and $s\geq 1$ . It is just (ii).
Consequently the proof of (ii) is complete.
References
[1] T.Ando and F.Hiai, Log majorization and complementary Golden-Thompson type inequalities, Linear Algebra Appl., 197, 198 (1994), 113-131.
[2] M.Fujii, Furuta’s inequality and its mean theoretic approach, J. Operator Theory, 23 (1990),
67-72.
[3] M.Fujii, T.Furuta and E.Kamei, Furuta’s inequality and its application to Ando’s theorem,
Linear Algebra Appl., 179 (1993), 161-169.
[4] M.Fujii and E.Kamei, Mean theoretic approach to the grand Furuta inequality, Proc. Amer.
Math. Soc., 124 (1996), 2751-2756.
[5] T.Furuta, A
$(1+2r)q\geq
0 assures $(B^{r}A^{p}Br)^{1/q}\geq B^{(p+2\Gamma})/q$ for r
p+2r$ , Proc. Amer. Math. Soc., 101 (1987), 85-88.
$\geq$
B
$\geq$
[6] T.Furuta, An elementary proof
(1989), 126.
of an
$\geq$
0, p
$\geq$
0, q
$\geq$
1 with
order preserving inequality, Proc. Japan Acad., 65
[7] T.Furuta, Two operator functions with monotone property, Proc. Amer. Math. Soc., 111
(1991), 511-516.
[8] T.Furuta, Applications
59 (1992), 180-190.
of order preserving
operator inequalities, Oper. Theory Adv. Appl.,
[9] T.Furuta, Extension of the Furuta inequality and Ando-Hiai
bra Appl., 19 (1995), 139-155.
$log$
-majorization, Linear Alge-
[10] T.Furuta and D.Wang, A decreasing operator function associated with the Furuta inequality,
Proc. Amer. Math. Soc., 126 (1998), 2427-2432
40
[11] T.Furuta, T.Yamazaki and M.Yanagida, Operator
inequality, Math. Inequal. Appl., 1 (1998), 123-130.
functions implying generalized Furuta
[12] T.Furuta, T.Yamazaki and M.Yanagida, Order preserving operator inequalities via Furuta
inequality, Math. Japon., 48 (1998), 471-476.
[13] T.Furuta, T.Yamazaki and M.Yanagida, Order preserving operator function via Furuta inB
equality “A
0 ensures
for p 1 and r 0”, to
appear in Proc. 96-IWOTA.
$(A^{\frac{r}{2}}A^{p}A \frac{r}{2})^{\frac{1+r}{p+r}}\geq(A^{\frac{r}{2}B^{p}}A^{\frac{r}{2})^{\frac{1+r}{p+r}}}$
$\geq$
$\geq$
$\geq$
$\geq$
[14] E.Kamei, A satellite to Furuta’s inequality, Math. Japon., 33 (1988), 883-886.
[15] K.Tanahashi, Best possibility
141-146.
[16] K.Tanahashi, Best possibility
Soc.
of the Furuta inequality, Proc. Amer. Math. Soc., 124 (1996),
of the grand Furuta inequality, to appear in Proc. Amer. Math.